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One liter of aqueus solution of suctrose `("molar mass"=324 g mol^(-1))` Weighing 1015 g is found g is foung to record an osmotic pressure of 4.82 atm at 293 K. What is the molality of the solution ?

Answer» Calculation of number of moles of solute.
`pi(nRT)/V or n=(piV)/(RT)`
`pi=4.82 atm, V=1L, R=0.0821 L atm K^(-1) mol^(-1),T=293 K`
`n=((4.82 atm)xx(1L))/((0.0821 L atm K^(-1)mol^(-1))xx(293K))=0.2 mol.`
Calculation of the molality of solution.
Mass of solute (sucrose)=`(0.2mol)xx(342 g mol^(-1))=68.4 g`
Mass of solution=1015 g
Mass of solvent(water)=1015-68.4=946.6 g =0.9466 kg
Molality of solution (m)= `(Mass of glucose//Gram molar mass)/(Mass of solvent in kg)
=((68.4g))/((342g mol^(-1)xx(0.9466 kg))=0.2113 mol//kg)= 0.2113 m`


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